### Hyperbola

• Find equation of hyperbola- regular and rectangular, eccentricity, directix, focus, latus rectum, asymptote, relation between a and b.

### Solved Example:

#### 1-6-01

Find the equation of the hyperbola having vertices: ($\pm$5,0), foci: ($\pm$8,0)

Solution:
For hyperbola, $\mathrm{vertices} = (\pm a,0) = (\pm5,0)$,
$\mathrm{foci} = (\pm c,0), c^2 = a^2 + b^2$,
$\mathrm{\ or,\ }b^2 = 8^2 - 5^2 = 64 - 25 = 39$
Equation of a hyperbola, $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 , \dfrac{x^2}{25} - \dfrac{y^2}{39} =1$

### Solved Example:

#### 1-6-02

What is the equation of the asymptote of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{4} =1$?

Solution:
$a = 3, b = 2$. Equation of the asymptote is: $y = \pm \dfrac{b}{a}x, y = \pm \dfrac{2}{3}x, 3y = 2x, 2x - 3y = 0$

### Solved Example:

#### 1-6-03

Find the equation of the hyperbola whose asymptotes are y = $\pm$ 2x and which passes through (5/2, 3).

Solution:
The equation of the asymptote is, $y = \dfrac{b}{a}x$.
Comparing with the given equation,$\dfrac{b}{a} = 2, b=2a$
So the equation of the hyperbola will become, $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{(2a)^{2}} = 1$
Substituting the given co-ordinates, $\dfrac{(5/2)^{2}}{a^{2}}-\dfrac{3^{2}}{(2a)^{2}} = 1, a = 2, b= 4$ $\dfrac{x^{2}}{2^{2}}-\dfrac{y^{2}}{4^{2}} = 1, 4x^2 - y^2 = 16$

### Solved Example:

#### 1-6-04

4$x^2$ - $y^2$ = 16 is the equation of a/an?

Solution:
$4x^{2}-y^{2}=16, \dfrac {x^{2}}{4}-\dfrac {y^{2}}{16}=1$ Because of the negative sign of $y^2$ term this is an equation of a hyperbola in the format, $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$

### Solved Example:

#### 1-6-05

Find the eccentricity of the curve: $9x^2 - 4y^2 - 36x + 8y = 4$

Solution:
\begin{align*} 9x^{2}-36x-4y^{2}+8y &=4\\ 9\left( x^{2}-4x+4\right) -36-4\left( y^{2}-2y+1\right) +4 &=4\\ \dfrac {9}{36}\left( x-2\right) ^{2}-\dfrac {4}{36}\left( y-1\right) ^{2} &=\dfrac {36}{36}\\ \dfrac {\left( x-2\right) ^{2}}{4}-\dfrac {\left( y-1\right) ^{2}}{9} &=1 \end{align*} a = 2, b =3 $b^{2} =a^{2}\left( e^{2}-1\right), 9 =4\left( e^{2}-1\right), 2.25 =e^{2}-1, e =1.80$

How far from the x-axis is the focus F of the hyperbola $x^2 - 2y^2 + 4x + 4y + 4 = 0?$
\begin{align*} x^{2}-2y^{2}+4x+4y+4 &=0\\ x^{2}+4x+4-2y^{2}+4y&=0\\ \left( x+2\right) ^{2}-2\left( y^{2}-2y\right) &=0\\ \left( x+2\right) ^{2}-2\left( y^{2}-2y+1\right) &=-2\\ \left( x+2\right) ^{2}-2\left( y-1\right) ^{2}&=-2\\ 2\left( y-1\right) ^{2}-\left( x+2\right) ^{2}&=2\\ \left( y-1\right) ^{2}-\dfrac {\left( x+2\right) ^{2}}{2}&=1\\ \end{align*} Comparing with, $\dfrac {Y^{2}}{a^{2}}-\dfrac {X^{2}}{b^{2}}=1, a=1, \ b=\sqrt {2}$ $X = x+2\ ,\ Y = y-1$ $e =\dfrac {\sqrt {a^{2}+b^{2}}}{a} =\dfrac {\sqrt {1+2}}{1} =\sqrt {3},$ $F\equiv \left( 0,\pm ae\right) =\left( 0,\pm \sqrt {3}\right)$ $X = x+2 \mathrm{\ Focus\ Shifted\ by}\ -2 \mathrm{\ in\ } x \mathrm{\ axis},$ $Y = y-1 \mathrm{\ Focus\ Shifted\ by}\ 1 \mathrm{\ in\ } y \mathrm{\ axis}$ $\mathrm{Focus\ Shifted\ by}=\left( -2,1\right)$ $F_{1} \equiv \left( -2,1+\sqrt {3}\right) =\left( -2,2.73\right),$ $F_{2}\equiv \left( -2,1-\sqrt {3}\right) =\left( -2,-0.73\right)$